Question

the standard deviation of the sampling distribution of the sample mean decreases when a) x increases b) n increases c) n decreases

Explanation:

Step1: Recall the formula for the standard deviation of the sampling distribution of the sample mean

The formula for the standard deviation of the sampling distribution of the sample mean (also known as the standard error) is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the population standard deviation and \(n\) is the sample size.

Step2: Analyze the relationship between \(n\) and \(\sigma_{\bar{x}}\)

From the formula \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\), we can see that when \(n\) (the sample size) increases, the value of \(\sqrt{n}\) increases. Since \(\sigma\) (assumed to be constant, as there is no information suggesting a change in the population standard deviation) is in the numerator and \(\sqrt{n}\) is in the denominator, the value of \(\sigma_{\bar{x}}\) (the standard deviation of the sampling distribution of the sample mean) decreases. For example, if \(\sigma = 10\), when \(n = 4\), \(\sigma_{\bar{x}}=\frac{10}{\sqrt{4}} = 5\); when \(n = 16\), \(\sigma_{\bar{x}}=\frac{10}{\sqrt{16}}= 2.5\).

Answer:

B. \(n\) increases